# What makes a limit 'go away'?

According to this video $$\lim\limits_{x\to\infty} \frac{11 - e^{-x}}{7} = \frac{11}{7}$$

I understand how this works, I don't understand the limit part though. I know $\lim\limits_{x\rightarrow \infty}{e^{-x}} = 0$, but what makes the limit disappear? I conclude that it disappears either because:

A. There are no more variables in the equation so no point to having a limit so we just start ignoring it

B. Is has been used once so it goes away.

Option B seems more likely, but it also it confusing because only one term seemed to be effected by the limit.

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To make the calculations simpler, look at $\frac{12-e^{-x}}{6}$. Use your calculator to compute this for $x=4$, $x=10$, $x=20$. – André Nicolas Feb 24 '14 at 16:49
I don't understand your difficulty. What makes the '+' go away in $1+1=2$? The fact that I did the computation! The same is true for the $\lim$ operator. – Marc van Leeuwen Feb 24 '14 at 20:51
@marcvanLeeuwen your answer: If you don't want to do addition then you have the option of subtracting 1 from both sides which makes the addition "go away". But I'm not sure if there is an inverse to Lim. – Back2Basics Feb 24 '14 at 22:23
$$\lim_{x\to ∞} f(x)+g(x) = \lim_{x\to ∞} f(x)+\lim_{x\to ∞}g(x)$$ when both limits exist. and $\lim_{x\to ∞} c = c$ so you can extract the $\frac{11}{7}$ from the limit and keep $\frac{e^x}{7}$ inside it. – ratchet freak Feb 25 '14 at 12:01

Option (A) is almost right!

Specifically, it is a property of limits that if $c$ is a constant, \begin{equation*} \lim_{x\to a}{c}=c. \end{equation*} You can prove this using the definition of a limit (that is, using a $\delta$-$\epsilon$ proof).

So $\lim_{x\to\infty}{\frac{11}{7}}=\frac{11}{7}$, having no limit on the right-hand side, just like $\lim_{x\to\infty}{e^{-x}}=0$, having no limit on the right-hand side.

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Indeed, and let me also point out for the OP's (or readers') sake that option (B) is actually wrong: the $\lim$ notation doesn't get "used up." That is, if $\lim g(x) = 0$, then $\lim f(x) + g(x)$ is still $\lim f(x)$, not plain old $f(x)$. (The specific chain of reasoning being $\lim[f(x) + g(x)] = \lim f(x) + \lim g(x) = \lim f(x) + 0 = \lim f(x)$.) – David Z Feb 24 '14 at 20:58
@DavidZ In layman's terms, it wasn't "used" in the first part of your example. It was distributed, then "used" on the g(x). So yeah, it's still kind of accurate. – Izkata Feb 24 '14 at 21:12
@Izkata well, based on the question I thought the OP might not know about the distributive property for limits, and perhaps his/her reasoning for coming up with option (B) was that when evaluating $\lim[f(x)+g(x)]$ you pick only one of the two terms to apply the limit to. I just wanted to make it explicit that that's not the case. – David Z Feb 24 '14 at 21:18
@DavidZ I'm hesitant even really to emphasize the distributive thing too much anyway. There are plenty of limits that only exist, where the limits of the summands by themselves don't exist. – Tim Seguine Feb 25 '14 at 10:36

The limit of a constant is just the value of constant, and when $\lim f(x)$ and $\lim g(x)$ both exist they satisfy $$\lim(f(x)+g(x)) = \lim f(x) + \lim g(x)$$ $$\lim(f(x)g(x)) = \lim f(x) \cdot \lim g(x)$$ In other words, here, you have $$\lim \frac{11-e^{-x}}{7} = \lim \frac{1}{7} \cdot \lim(11-e^x) = \lim \frac{1}{7}(\lim 11 - \lim e^{-x})$$ Since $\frac{1}{7}$ and $11$ are constant, $\lim e^{-x} = 0$, you get $$\lim \frac{11-e^{-x}}{7} = \frac{1}{7} \cdot (11 + 0) = \frac{11}{7}$$

(I've left the subscript $x \to \infty$ off the limit signs.)

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Note this is true only when $\lim(f(x))$ and $\lim(g(x))$ are finite. – perfectionist Feb 24 '14 at 23:29
@perfectionist: I take 'exists' to mean 'is defined and is a real number', so that's covered. Perhaps I should have made that clearer in my answer, but I think this convention is common. – Clive Newstead Feb 24 '14 at 23:52

If we continuously extend the real exponential to have values $e^{+\infty} = +\infty$ and $e^{-\infty} = 0$, then this can be seen as taking the limit of a continuous function, which can be done by simply plugging in the limiting value:

$$\lim_{x\to +\infty} \frac{11 - e^{-x}}{7} = \frac{11 - e^{-(+\infty)}}{7} = \frac{11 - e^{-\infty}}{7} = \frac{11 - 0}{7} = \frac{11}{7}$$

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I guess I like this, but it's probably more clear if you first make the substitution $\mathrm{e}^{-x}\Rightarrow\frac{1}{\mathrm{e}^{x}}$ instead of rewriting $-(+\infty)\Rightarrow-\infty$. – NikolajK Dec 12 '14 at 13:33

I can guess that your doubt comes from a first step that you do in your mind that goes like this: "$\lim_{x\rightarrow \infty}{e^{-x}} = 0$, so I can replace $e^{-x}$ with $0$ in the text and I remain with $$\lim_{x\to ∞} \frac{11-0}{7},$$ then what happens to the limit in this last equation?"

This reflects a common misunderstanding. Repeat after me: in a limit, one cannot replace arbitrary subexpressions with their limit. There is no theorem that says that. If you've been doing that, you are wrong; change your habits.

For an easier example when this doesn't work, $\lim_{x\to 0} x = 0$, but $$\lim_{x\to 0} \frac{x+x^3}{x^2} \neq \lim_{x\to 0} \frac{0+x^3}{x^2}.$$

Clive Newstead's and Ilmari Karonen's answers show some alternative manipulations that are legitimate by theorems.

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As EPAstor notes, your reason (A) is closer to the truth, but it's not the complete answer.

The reason why $\displaystyle \lim_{x \to \infty} e^{-x} = 0$ implies $\displaystyle \lim_{x \to \infty} \frac{11 - e^{-x}}{7} = \frac{11}{7}$ is that

$$\lim_{x \to \infty} \frac{11 - e^{-x}}{7} = \frac{\lim_{x \to \infty} 11 - e^{-x}}{7} = \frac{11 - (\lim_{x \to \infty} e^{-x})}{7} = \frac{11 - 0}{7} = \frac{11}{7}$$

and the reason we can move the limit inside the fraction like that is that the functions $$z \mapsto 11 - z \quad \text{and} \quad z \mapsto \frac z 7$$ are both everywhere continuous. It is a rather straightforward consequence of the definition of continuity (prove it!) that, if a function $f$ is continuous at $c = \displaystyle\lim_{x \to a} z(x)$, then $$\lim_{x \to a} f(z(x)) = f(\lim_{x \to a} z(x)) = f(c).$$

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First note that $$\lim\limits_{x\to\infty} f(x)= f\left(\lim\limits_{x\to\infty} x\right)$$ And $$\lim\limits_{x\to\infty} e^{-x}= \lim\limits_{x\to\infty} \frac{1}{e^{x}} =0$$ Therefore $$\lim\limits_{x\to\infty} \frac{11 - e^{-x}}{7} = \frac{11 - \lim\limits_{x\to\infty} e^{-x}}{7} =\frac{11}{7}$$

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